1 / 2

MOLES = MASS/GFM Al

PROBLEM 4 OF LIMITING REAGENT SET. 4 Al + 3 O 2  2 Al 2 O 3 FIRST DETERMINE IF THERE IS A LIMITING REAGENT . FIND MOLES OF ALUMINUM METAL FIND MOLES OF OXYGEN GAS. MOLES = MASS/GFM Al. MOLES = 2.5g Al/ 26.98g/mol. MOLES = 0.092661mol Al. MOLES = MASS/GFM O 2.

melissan
Télécharger la présentation

MOLES = MASS/GFM Al

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. PROBLEM 4 OF LIMITING REAGENT SET.4 Al + 3O2 2Al2O3 FIRST DETERMINE IF THERE IS A LIMITING REAGENT.FIND MOLES OF ALUMINUM METAL FIND MOLES OF OXYGEN GAS. MOLES = MASS/GFM Al MOLES = 2.5g Al/ 26.98g/mol MOLES = 0.092661mol Al MOLES = MASS/GFM O2 MOLES = 2.5g O2 / 32.00g/mol MOLES = 0.0781mol O2 RATIO BOTH REACTANTS TO FIND LIMITING REAGENT. O23 = 0.07810 mol O2 Al 4 0.09266 mol Al 0.75 = .842 The actual ratio is larger than the theoretical, therefore the actual ratio fraction is too large, the denominator is to small Al is limiting.

  2. 4 Al + 3O2 2Al2O3 STEP 1 Identify known (the substance that can be converted to moles), choose the correct equation to convert to moles. MOLES = MASS/GFM Al MOLES = 2.5g Al/ 26.98g/mol MOLES = 0.092661mol Al STEP 2 RATIO known to objective, Al=4=0.092661 , Al2O32XX = 0.0463. mol Al2O3 STEP 3 Convert Objective to required units. % yield = experimental/theoretical 3.5 / 4.7236 =74% MOLES = MASS/GFM Al2O3 .0463 mol= mass/ 101.96 Mass =4.7236 g Al2O3

More Related